How to show $\,2^n \equiv 8\pmod{24}\,$ for odd $\,n\ge 3$?

Question

Show that if $n\geq3$ is odd, then $2^n-1\equiv7\mod24$.

I tried solving this backwardly. We want to prove that $2^3(2^{n-3}-1)=2^n-2^3\equiv0\mod24$. Since $\frac{24}{2^3}=3$, this leaves us to prove that $3\mid2^{n-3}-1$. From here on I got stuck, although I feel like I am ignoring something very simple.

Once the result is proved, I need to find all $n\in\mathbb{Z}_{>0}$ for which $2^n-1$ divides $3^n-1$.

This problem is given in a chapter on the Legendre symbol, introducing the law of quadratic reciprocity.

Answer 1

Remember $n$ is odd, so writing $n=2k+1$ gives $$2^{n-3}-1\equiv 2^{2k-2}-1\equiv (2^2)^{k-1}-1\equiv 1-1\equiv 0\pmod{3}$$

Answer 2

Hint $\,\ 2^{\overbrace{\large 3+2k}^{\LARGE n}}\bmod 24\, =\, 2^{\large 3}(\overbrace{\color{#c00}2^{\large\color{#c00} 2k}\!\bmod 3}^{\large \color{#c00}{2^{\LARGE 2}}\ \equiv\ 1})\, =\, 2^{\large 3}\ $

We used: $\,\ ab\bmod ac\, =\, a(b\bmod c)\,= $ mod Distributive Law to factor $\,a = 2^{\large 3}$ out of the mod.

Answer 3

In one hand, since $n\geq 3$ then $8|2^n$, i.e. $2^n\equiv 0\bmod 8$. Hence $$2^n-1\equiv 7\mod 8$$

On the other hand, since $n$ is odd, we have $2^n\equiv(-1)^n=-1\bmod 3$. Hence, $2^n-1\equiv -2\bmod 3$. In particular $$2^n-1\equiv 7\mod 3.$$

So, we conclude $2^n-1\equiv 7\mod 24.$